\(\int \frac {1}{\sqrt [4]{3+b x} (c+b x)^{5/4}} \, dx\) [651]

Optimal result

Integrand size = 19, antiderivative size = 71 \[ \int \frac {1}{\sqrt [4]{3+b x} (c+b x)^{5/4}} \, dx=-\frac {4 \sqrt [4]{3+b x} \sqrt [4]{\frac {c+b x}{3+b x}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {-3+c}}{\sqrt {3+b x}}\right )\right |2\right )}{b \sqrt {-3+c} \sqrt [4]{c+b x}} \] Output:

-4*(b*x+3)^(1/4)*((b*x+c)/(b*x+3))^(1/4)*EllipticE(sin(1/2*arctan((-3+c)^( 
1/2)/(b*x+3)^(1/2))),2^(1/2))/b/(-3+c)^(1/2)/(b*x+c)^(1/4)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.03 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.89 \[ \int \frac {1}{\sqrt [4]{3+b x} (c+b x)^{5/4}} \, dx=\frac {4 (3+b x)^{3/4} \left (\frac {c+b x}{-3+c}\right )^{5/4} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{4},\frac {7}{4},\frac {3+b x}{3-c}\right )}{3 b (c+b x)^{5/4}} \] Input:

Integrate[1/((3 + b*x)^(1/4)*(c + b*x)^(5/4)),x]
 

Output:

(4*(3 + b*x)^(3/4)*((c + b*x)/(-3 + c))^(5/4)*Hypergeometric2F1[3/4, 5/4, 
7/4, (3 + b*x)/(3 - c)])/(3*b*(c + b*x)^(5/4))
 

Rubi [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(143\) vs. \(2(71)=142\).

Time = 0.27 (sec) , antiderivative size = 143, normalized size of antiderivative = 2.01, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {61, 73, 840, 842, 858, 807, 226}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {1}{\sqrt [4]{b x+3} (b x+c)^{5/4}} \, dx\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {2 \int \frac {1}{\sqrt [4]{b x+3} \sqrt [4]{c+b x}}dx}{3-c}-\frac {4 (b x+3)^{3/4}}{b (3-c) \sqrt [4]{b x+c}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {8 \int \frac {\sqrt {b x+3}}{\sqrt [4]{c+b x}}d\sqrt [4]{b x+3}}{b (3-c)}-\frac {4 (b x+3)^{3/4}}{b (3-c) \sqrt [4]{b x+c}}\)

\(\Big \downarrow \) 840

\(\displaystyle \frac {8 \left (\frac {(b x+c)^{3/4}}{2 \sqrt [4]{b x+3}}-\frac {1}{2} (3-c) \int \frac {1}{\sqrt {b x+3} \sqrt [4]{c+b x}}d\sqrt [4]{b x+3}\right )}{b (3-c)}-\frac {4 (b x+3)^{3/4}}{b (3-c) \sqrt [4]{b x+c}}\)

\(\Big \downarrow \) 842

\(\displaystyle \frac {8 \left (\frac {(b x+c)^{3/4}}{2 \sqrt [4]{b x+3}}-\frac {(3-c) \sqrt [4]{b x+3} \sqrt [4]{1-\frac {3-c}{b x+3}} \int \frac {1}{(b x+3)^{3/4} \sqrt [4]{1-\frac {3-c}{b x+3}}}d\sqrt [4]{b x+3}}{2 \sqrt [4]{b x+c}}\right )}{b (3-c)}-\frac {4 (b x+3)^{3/4}}{b (3-c) \sqrt [4]{b x+c}}\)

\(\Big \downarrow \) 858

\(\displaystyle \frac {8 \left (\frac {(3-c) \sqrt [4]{b x+3} \sqrt [4]{1-\frac {3-c}{b x+3}} \int \frac {1}{\sqrt [4]{b x+3} \sqrt [4]{(c-3) (b x+3)+1}}d\frac {1}{\sqrt [4]{b x+3}}}{2 \sqrt [4]{b x+c}}+\frac {(b x+c)^{3/4}}{2 \sqrt [4]{b x+3}}\right )}{b (3-c)}-\frac {4 (b x+3)^{3/4}}{b (3-c) \sqrt [4]{b x+c}}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {8 \left (\frac {(3-c) \sqrt [4]{b x+3} \sqrt [4]{1-\frac {3-c}{b x+3}} \int \frac {1}{\sqrt [4]{1-(3-c) \sqrt {b x+3}}}d\sqrt {b x+3}}{4 \sqrt [4]{b x+c}}+\frac {(b x+c)^{3/4}}{2 \sqrt [4]{b x+3}}\right )}{b (3-c)}-\frac {4 (b x+3)^{3/4}}{b (3-c) \sqrt [4]{b x+c}}\)

\(\Big \downarrow \) 226

\(\displaystyle \frac {8 \left (\frac {\sqrt {3-c} \sqrt [4]{b x+3} \sqrt [4]{1-\frac {3-c}{b x+3}} E\left (\left .\frac {1}{2} \arcsin \left (\sqrt {3-c} \sqrt {b x+3}\right )\right |2\right )}{2 \sqrt [4]{b x+c}}+\frac {(b x+c)^{3/4}}{2 \sqrt [4]{b x+3}}\right )}{b (3-c)}-\frac {4 (b x+3)^{3/4}}{b (3-c) \sqrt [4]{b x+c}}\)

Input:

Int[1/((3 + b*x)^(1/4)*(c + b*x)^(5/4)),x]
 

Output:

(-4*(3 + b*x)^(3/4))/(b*(3 - c)*(c + b*x)^(1/4)) + (8*((c + b*x)^(3/4)/(2* 
(3 + b*x)^(1/4)) + (Sqrt[3 - c]*(3 + b*x)^(1/4)*(1 - (3 - c)/(3 + b*x))^(1 
/4)*EllipticE[ArcSin[Sqrt[3 - c]*Sqrt[3 + b*x]]/2, 2])/(2*(c + b*x)^(1/4)) 
))/(b*(3 - c))
 

Defintions of rubi rules used

rule 61

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d 
, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 226

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(2/(a^(1/4)*Rt[-b/a, 2] 
))*EllipticE[(1/2)*ArcSin[Rt[-b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ 
[a, 0] && NegQ[b/a]
 

rule 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

rule 840

Int[(x_)^2/((a_) + (b_.)*(x_)^4)^(1/4), x_Symbol] :> Simp[(a + b*x^4)^(3/4) 
/(2*b*x), x] + Simp[a/(2*b)   Int[1/(x^2*(a + b*x^4)^(1/4)), x], x] /; Free 
Q[{a, b}, x] && NegQ[b/a]
 

rule 842

Int[1/((x_)^2*((a_) + (b_.)*(x_)^4)^(1/4)), x_Symbol] :> Simp[x*((1 + a/(b* 
x^4))^(1/4)/(a + b*x^4)^(1/4))   Int[1/(x^3*(1 + a/(b*x^4))^(1/4)), x], x] 
/; FreeQ[{a, b}, x] && NegQ[b/a]
 

rule 858

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + 
b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int 
egerQ[m]
 

Maple [F]

\[\int \frac {1}{\left (b x +3\right )^{\frac {1}{4}} \left (b x +c \right )^{\frac {5}{4}}}d x\]

Input:

int(1/(b*x+3)^(1/4)/(b*x+c)^(5/4),x)
 

Output:

int(1/(b*x+3)^(1/4)/(b*x+c)^(5/4),x)
 

Fricas [F]

\[ \int \frac {1}{\sqrt [4]{3+b x} (c+b x)^{5/4}} \, dx=\int { \frac {1}{{\left (b x + c\right )}^{\frac {5}{4}} {\left (b x + 3\right )}^{\frac {1}{4}}} \,d x } \] Input:

integrate(1/(b*x+3)^(1/4)/(b*x+c)^(5/4),x, algorithm="fricas")
 

Output:

integral((b*x + c)^(3/4)*(b*x + 3)^(3/4)/(b^3*x^3 + (2*b^2*c + 3*b^2)*x^2 
+ 3*c^2 + (b*c^2 + 6*b*c)*x), x)
 

Sympy [F]

\[ \int \frac {1}{\sqrt [4]{3+b x} (c+b x)^{5/4}} \, dx=\int \frac {1}{\sqrt [4]{b x + 3} \left (b x + c\right )^{\frac {5}{4}}}\, dx \] Input:

integrate(1/(b*x+3)**(1/4)/(b*x+c)**(5/4),x)
 

Output:

Integral(1/((b*x + 3)**(1/4)*(b*x + c)**(5/4)), x)
 

Maxima [F]

\[ \int \frac {1}{\sqrt [4]{3+b x} (c+b x)^{5/4}} \, dx=\int { \frac {1}{{\left (b x + c\right )}^{\frac {5}{4}} {\left (b x + 3\right )}^{\frac {1}{4}}} \,d x } \] Input:

integrate(1/(b*x+3)^(1/4)/(b*x+c)^(5/4),x, algorithm="maxima")
 

Output:

integrate(1/((b*x + c)^(5/4)*(b*x + 3)^(1/4)), x)
 

Giac [F]

\[ \int \frac {1}{\sqrt [4]{3+b x} (c+b x)^{5/4}} \, dx=\int { \frac {1}{{\left (b x + c\right )}^{\frac {5}{4}} {\left (b x + 3\right )}^{\frac {1}{4}}} \,d x } \] Input:

integrate(1/(b*x+3)^(1/4)/(b*x+c)^(5/4),x, algorithm="giac")
 

Output:

integrate(1/((b*x + c)^(5/4)*(b*x + 3)^(1/4)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt [4]{3+b x} (c+b x)^{5/4}} \, dx=\int \frac {1}{{\left (b\,x+3\right )}^{1/4}\,{\left (c+b\,x\right )}^{5/4}} \,d x \] Input:

int(1/((b*x + 3)^(1/4)*(c + b*x)^(5/4)),x)
 

Output:

int(1/((b*x + 3)^(1/4)*(c + b*x)^(5/4)), x)
 

Reduce [F]

\[ \int \frac {1}{\sqrt [4]{3+b x} (c+b x)^{5/4}} \, dx=\int \frac {1}{\left (b x +3\right )^{\frac {1}{4}} \left (b x +c \right )^{\frac {1}{4}} b x +\left (b x +3\right )^{\frac {1}{4}} \left (b x +c \right )^{\frac {1}{4}} c}d x \] Input:

int(1/(b*x+3)^(1/4)/(b*x+c)^(5/4),x)
 

Output:

int(1/((b*x + 3)**(1/4)*(b*x + c)**(1/4)*b*x + (b*x + 3)**(1/4)*(b*x + c)* 
*(1/4)*c),x)